Question: $\dfrac{ d + 9e }{ 3 } = \dfrac{ 8d - 3f }{ -9 }$ Solve for $d$.
Multiply both sides by the left denominator. $\dfrac{ d + 9e }{ {3} } = \dfrac{ 8d - 3f }{ -9 }$ ${3} \cdot \dfrac{ d + 9e }{ {3} } = {3} \cdot \dfrac{ 8d - 3f }{ -9 }$ $d + 9e = {3} \cdot \dfrac { 8d - 3f }{ -9 }$ Multiply both sides by the right denominator. $d + 9e = 3 \cdot \dfrac{ 8d - 3f }{ -{9} }$ $-{9} \cdot \left( d + 9e \right) = -{9} \cdot 3 \cdot \dfrac{ 8d - 3f }{ -{9} }$ $-{9} \cdot \left( d + 9e \right) = 3 \cdot \left( 8d - 3f \right)$ Distribute both sides $-{9} \cdot \left( d + 9e \right) = {3} \cdot \left( 8d - 3f \right)$ $-{9}d - {81}e = {24}d - {9}f$ Combine $d$ terms on the left. $-{9d} - 81e = {24d} - 9f$ $-{33d} - 81e = -9f$ Move the $e$ term to the right. $-33d - {81e} = -9f$ $-33d = -9f + {81e}$ Isolate $d$ by dividing both sides by its coefficient. $-{33}d = -9f + 81e$ $d = \dfrac{ -9f + 81e }{ -{33} }$ All of these terms are divisible by $3$ Divide by the common factor and swap signs so the denominator isn't negative. $d = \dfrac{ {3}f - {27}e }{ {11} }$